Circle

A circle is a round, two-dimensional shape. All points on the edge of the circle are at the same distance from the center.

A Circle

The point on the side. Mathematicians use the letter $r$ for the length of a circle's radius. The center of a circle is the point in the very middle. It is often written as $O$.

The diameter (meaning "all the way across") of a circle is a straight line that goes from one side to the opposite and right through the center of the circle. Mathematicians use the letter $d$ for the length of this line. The diameter of a circle is equal to twice its radius ($d$ equals $2$ times $r$):

$d=2r$

The circumference (meaning "all the way around") of a circle is the line that goes around the center of the circle. Mathematicians use the letter $c$ for the length of this line.

The number $\pi$ (written as the Greek letter pi) is a very useful number. It is the length of the circumference divided by the length of the diameter ($\pi$ equals $c$ divided by $d$). As a fraction the number $\pi$ is equal to about ${\frac {22}{7}}$ or ${\frac {355}{113}}$ (which is closer) and as a number it is about $3.1415926536$.

	$\pi ={\frac {c}{d}}$
$\therefore {\textrm {(therefore)}}$	$c=2\pi r$
	$c=\pi d$

The area of the circle is equal to $\pi$ times the area of the gray square.

The area, $A$, inside a circle is equal to the radius multiplied by itself, then multiplied by $\pi$ ($A$ equals $\pi$ times $r$ times $r$).

$A=\pi r^{2}$

Calculating π

$\pi$ can be measured by drawing a circle, then measuring its diameter ($d$) and circumference ($c$). This is because the circumference of a circle is always equal to $\pi$ times its diameter.

$\pi ={\frac {c}{d}}$

$\pi$ can also be calculated by only using mathematical methods. Most methods used for calculating the value of $\pi$ have desirable mathematical properties. However, they are hard to understand without knowing trigonometry and calculus. However, some methods are quite simple, such as this form of the Gregory-Leibniz series:

$\pi ={\frac {4}{1}}-{\frac {4}{3}}+{\frac {4}{5}}-{\frac {4}{7}}+{\frac {4}{9}}-{\frac {4}{11}}+\cdots$

While that series is easy to write and calculate, it is not easy to see why it equals $\pi$. A much easier way to approach is to draw an imaginary circle of radius $r$ centered at the origin. Then any point $(x,y)$ whose distance $d$ from the origin is less than $r$, calculated by the Pythagorean theorem, will be inside the circle:

$d={\sqrt {x^{2}+y^{2}}}$

Finding a set of points inside the circle allows the circle's area $A$ to be estimated, for example, by using integer coordinates for a big $r$. Since the area $A$ of a circle is $\pi$ times the radius squared, $\pi$ can be approximated by using the following formula:

$\pi ={\frac {A}{r^{2}}}$

Calculating measures of a circle

Area

Using the radius: $A=\pi r^{2}={\frac {\tau r^{2}}{2}}$

Using the diameter: $A={\frac {\pi d^{2}}{4}}={\frac {\tau d^{2}}{8}}$

Using the circumference: $A={\frac {c^{2}}{2\tau }}={\frac {c^{2}}{4\pi }}$

Circumference

Using the radius: $c=\tau r=2\pi r$

Using the diameter: $c=\pi d={\frac {\tau d}{2}}$

Using the area: $c={\sqrt {2\tau A}}=2{\sqrt {\pi A}}$

Diameter

Using the radius: $d=2r$

Using the circumference: $d={\frac {c}{\pi }}={\frac {2c}{\tau }}$

Using the area: $d=2{\sqrt {\frac {A}{\pi }}}=2{\sqrt {\frac {2A}{\tau }}}$

Radius

Using the diameter: $r={\frac {d}{2}}$

Using the circumference: $r={\frac {c}{\tau }}={\frac {c}{2\pi }}$

Using the area: $r={\sqrt {\frac {A}{\pi }}}={\sqrt {\frac {2A}{\tau }}}$

Equations

Cartesian coordinates

The equation of a circle in Cartesian coordinates, with the centre at $(h,k)$ and radius $r$, is given by $$(x-h)^{2}+(y-k)^{2}=r^{2}$$

Polar coordinates

A circle centred at the origin with radius $R$ is the equivalent of a constant function, $$r(\theta )=R$$

A circle with a centre at a different location $(h,k)$ and radius $R$ can be found by using the conversion between polar and Cartesian coordinates: $${\begin{aligned}x={}&r\cos \theta \\y={}&r\sin \theta \end{aligned}}$$ Substituting this into the equation for a circle in rectangular coordinates gives $${\begin{aligned}(r\cos \theta -h)^{2}+(r\sin \theta -k)^{2}={}&R^{2}\\r^{2}\cos ^{2}\theta -2rh\cos \theta +h^{2}+r^{2}\sin ^{2}\theta -2rk\sin \theta +k^{2}={}&R^{2}\\r^{2}-2(h\cos \theta +k\sin \theta )r+h^{2}+k^{2}-R^{2}={}&0\end{aligned}}$$ This is a quadratic equation in $r$. The quadratic formula can be used: $${\begin{aligned}a={}&1\\b={}&-2(h\cos \theta +k\sin \theta )\\c={}&h^{2}+k^{2}-R^{2}\\r={}&{\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\\={}&{\frac {2(h\cos \theta +k\sin \theta )\pm {\sqrt {4(h\cos \theta +k\sin \theta )^{2}-4(h^{2}+k^{2}-R^{2})}}}{2}}\\={}&h\cos \theta +k\sin \theta \pm {\sqrt {h^{2}\cos ^{2}\theta +2hk\cos \theta \sin \theta +k^{2}\sin ^{2}\theta -h^{2}-k^{2}+R^{2}}}\\={}&h\cos \theta +k\sin \theta \pm {\sqrt {R^{2}-h^{2}\sin ^{2}\theta +2hk\cos \theta \sin \theta -k^{2}\cos ^{2}\theta }}\\={}&h\cos \theta +k\sin \theta \pm {\sqrt {R^{2}-(h\sin \theta -k\cos \theta )^{2}}}\end{aligned}}$$ Whether the square root is added or subtracted does not matter as long as $\theta$ is allowed to make a full rotation: ${\begin{aligned}r(\theta +\pi )={}&h\cos(\theta +\pi )+k\sin(\theta +\pi )+{\sqrt {R^{2}-(h\sin(\theta +\pi )-k\cos(\theta +\pi ))^{2}}}\\={}&-h\cos \theta -k\sin \theta +{\sqrt {R^{2}-(-h\sin \theta +k\cos \theta )^{2}}}\\={}&-(h\cos \theta +k\sin \theta -{\sqrt {R^{2}-(h\sin \theta -k\cos \theta )^{2}}})\end{aligned}}$ In polar coordinates, the points $(\theta ,r)$ and $(\theta +\pi ,-r)$ are equal, so this means that both the positive and negative conjugates are included by other half of the domain.

Related pages

• Concentric
• Semicircle
• Sphere
• Squaring the circle
• Pi
• Tau

Other websites

• Calculate the measures of a circle online